Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 62128 by maxmathsup by imad last updated on 15/Jun/19

let U_n = ∫_0 ^∞ ((cos(nx))/((x^2 +n^2 )^3 ))dx with n≥1 1) calculate U_n intrems of n 2) find lim_(n→+∞) n U_n 3) calculate lim_(n→+∞) n^2 U_n 4) study the convervence of U_n

$${let}\:{U}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({nx}\right)}{\left({x}^{\mathrm{2}} \:+{n}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx}\:\:{with}\:{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{U}_{{n}} \:{intrems}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} {n}\:{U}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{lim}_{{n}\rightarrow+\infty} {n}^{\mathrm{2}} \:{U}_{{n}} \\ $$$$\left.\mathrm{4}\right)\:{study}\:{the}\:{convervence}\:{of}\:{U}_{{n}} \\ $$

Commented by maxmathsup by imad last updated on 16/Jun/19

1) U_n =∫_0 ^∞ ((cos(nx))/((x^2 +n^2 )^3 ))dx changement x =nt give U_n =∫_0 ^∞ ((cos(n^2 t))/((n^2 t^2 +n^2 )^3 )) ndt =(1/n^5 ) ∫_0 ^∞ ((cos(n^2 t))/((t^2 +1)^3 )) dt ⇒2n^5 U_n =∫_(−∞) ^(+∞) ((cos(n^2 t))/((t^2 +1)^3 ))dt =Re(∫_(−∞) ^(+∞) (e^(in^2 t) /((t^2 +1)^3 ))dt) let consider thecomplex function ϕ(z)=(e^(in^2 z) /((z^2 +1)^3 )) we have ϕ(z)=(e^(inz^2 ) /((z−i)^3 (z+i)^3 )) the poles of ϕ are +^− i (triples) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((3−1)!)){(z−i)^3 ϕ(z)}^((2)) =lim_(z→i) (1/2){ (e^(in^2 z) /((z+i)^3 ))}^((2)) but { (e^(in^2 z) /((z+i)^3 ))}^((2)) ={((in^2 e^(in^2 z) (z+i)^3 −3(z+i)^2 e^(in^2 z) )/((z+i)^6 ))}^((1)) ={(((in^2 (z+i)−3)e^(in^2 z) )/((z+i)^4 ))}^((1)) ={(((in^2 z−n^2 −3)e^(in^2 z) )/((z+i)^4 ))}^((1)) =(((in^2 +in^2 (in^2 z−n^2 −3))e^(in^2 z) (z+i)^4 −4(z+i)^3 (in^2 z−n^2 −3)e^(in^2 z) )/((z+i)^8 )) =(({(in^2 −n^4 z −in^4 −3in^2 )(z+i)−4in^2 z+4n^2 +12}e^(in^2 z) )/((z+i)^5 )) ⇒ Res(ϕ,i) =(1/2) (({ 2i(in^2 −n^4 i−in^4 −3in^2 ) +4n^2 +4n^2 +12}e^(−n^2 ) )/((2i)^5 )) =(1/2^6 ) (({−2n^2 +2n^4 +2n^4 +6n^2 +8n^2 +12}e^(−n^2 ) )/i) =(1/(i 2^6 )){12n^2 +4n^4 +12}e^(−n^2 ) =(4/(i 2^6 )){ 3n^2 +n^4 +3}e^(−n^2 ) = (1/(16i)){n^4 +3n^2 +3} e^(−n^2 ) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/(16i)){ n^4 +3n^2 +3}e^(−n^2 ) =(π/8)(n^4 +3n^2 +3)e^(−n^2 ) =2n^5 U_n ⇒ U_n =(π/(16n^5 ))(n^4 +3n^2 +3)e^(−n^2 ) 3) n^2 U_n =(π/(16n^3 ))(n^4 +3n^2 +3)e^(−n^2 ) =(π/(16))(n +3 +(3/n^3 ))e^(−n^2 ) ⇒ lim_(n→+∞) n^2 U_n =lim_(n→+∞) (π/(16))(n+3)e^(−n^2 ) =0

$$\left.\mathrm{1}\right)\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left({nx}\right)}{\left({x}^{\mathrm{2}} \:+{n}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx}\:\:{changement}\:{x}\:={nt}\:{give} \\ $$$${U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({n}^{\mathrm{2}} {t}\right)}{\left({n}^{\mathrm{2}} {t}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)^{\mathrm{3}} }\:{ndt}\:=\frac{\mathrm{1}}{{n}^{\mathrm{5}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left({n}^{\mathrm{2}} {t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:{dt}\:\Rightarrow\mathrm{2}{n}^{\mathrm{5}} \:{U}_{{n}} =\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({n}^{\mathrm{2}} {t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }{dt} \\ $$$$={Re}\left(\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{in}^{\mathrm{2}} {t}} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }{dt}\right)\:\:{let}\:{consider}\:{thecomplex}\:{function}\:\varphi\left({z}\right)=\frac{{e}^{{in}^{\mathrm{2}} {z}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${we}\:{have}\:\varphi\left({z}\right)=\frac{{e}^{{inz}^{\mathrm{2}} } }{\left({z}−{i}\right)^{\mathrm{3}} \left({z}+{i}\right)^{\mathrm{3}} }\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{i}\:\left({triples}\right)\:{residus}\:{theorem} \\ $$$${give}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{3}} \varphi\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\frac{{e}^{{in}^{\mathrm{2}} {z}} }{\left({z}+{i}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \:\:{but} \\ $$$$\left\{\:\frac{{e}^{{in}^{\mathrm{2}} {z}} }{\left({z}+{i}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \:=\left\{\frac{{in}^{\mathrm{2}} {e}^{{in}^{\mathrm{2}} {z}} \left({z}+{i}\right)^{\mathrm{3}} −\mathrm{3}\left({z}+{i}\right)^{\mathrm{2}} {e}^{{in}^{\mathrm{2}} {z}} }{\left({z}+{i}\right)^{\mathrm{6}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\left\{\frac{\left({in}^{\mathrm{2}} \left({z}+{i}\right)−\mathrm{3}\right){e}^{{in}^{\mathrm{2}} {z}} }{\left({z}+{i}\right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \:=\left\{\frac{\left({in}^{\mathrm{2}} {z}−{n}^{\mathrm{2}} −\mathrm{3}\right){e}^{{in}^{\mathrm{2}} {z}} }{\left({z}+{i}\right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\left({in}^{\mathrm{2}} \:+{in}^{\mathrm{2}} \left({in}^{\mathrm{2}} {z}−{n}^{\mathrm{2}} −\mathrm{3}\right)\right){e}^{{in}^{\mathrm{2}} {z}} \left({z}+{i}\right)^{\mathrm{4}} \:−\mathrm{4}\left({z}+{i}\right)^{\mathrm{3}} \left({in}^{\mathrm{2}} {z}−{n}^{\mathrm{2}} −\mathrm{3}\right){e}^{{in}^{\mathrm{2}} {z}} }{\left({z}+{i}\right)^{\mathrm{8}} } \\ $$$$=\frac{\left\{\left({in}^{\mathrm{2}} −{n}^{\mathrm{4}} {z}\:−{in}^{\mathrm{4}} −\mathrm{3}{in}^{\mathrm{2}} \right)\left({z}+{i}\right)−\mathrm{4}{in}^{\mathrm{2}} {z}+\mathrm{4}{n}^{\mathrm{2}} \:+\mathrm{12}\right\}{e}^{{in}^{\mathrm{2}} {z}} }{\left({z}+{i}\right)^{\mathrm{5}} }\:\Rightarrow \\ $$$${Res}\left(\varphi,{i}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\left\{\:\mathrm{2}{i}\left({in}^{\mathrm{2}} −{n}^{\mathrm{4}} {i}−{in}^{\mathrm{4}} −\mathrm{3}{in}^{\mathrm{2}} \right)\:+\mathrm{4}{n}^{\mathrm{2}} \:+\mathrm{4}{n}^{\mathrm{2}} \:+\mathrm{12}\right\}{e}^{−{n}^{\mathrm{2}} } }{\left(\mathrm{2}{i}\right)^{\mathrm{5}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\:\frac{\left\{−\mathrm{2}{n}^{\mathrm{2}} +\mathrm{2}{n}^{\mathrm{4}} +\mathrm{2}{n}^{\mathrm{4}} \:+\mathrm{6}{n}^{\mathrm{2}} \:+\mathrm{8}{n}^{\mathrm{2}} \:+\mathrm{12}\right\}{e}^{−{n}^{\mathrm{2}} } }{{i}}\:=\frac{\mathrm{1}}{{i}\:\mathrm{2}^{\mathrm{6}} }\left\{\mathrm{12}{n}^{\mathrm{2}} \:+\mathrm{4}{n}^{\mathrm{4}} \:+\mathrm{12}\right\}{e}^{−{n}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}}{{i}\:\mathrm{2}^{\mathrm{6}} }\left\{\:\mathrm{3}{n}^{\mathrm{2}} \:+{n}^{\mathrm{4}} \:+\mathrm{3}\right\}{e}^{−{n}^{\mathrm{2}} } \:=\:\frac{\mathrm{1}}{\mathrm{16}{i}}\left\{{n}^{\mathrm{4}} \:+\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{3}\right\}\:{e}^{−{n}^{\mathrm{2}} \:} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\mathrm{2}{i}\pi}{\mathrm{16}{i}}\left\{\:{n}^{\mathrm{4}} \:+\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{3}\right\}{e}^{−{n}^{\mathrm{2}} } \:=\frac{\pi}{\mathrm{8}}\left({n}^{\mathrm{4}} +\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{3}\right){e}^{−{n}^{\mathrm{2}} } \:=\mathrm{2}{n}^{\mathrm{5}} \:{U}_{{n}} \:\Rightarrow \\ $$$${U}_{{n}} =\frac{\pi}{\mathrm{16}{n}^{\mathrm{5}} }\left({n}^{\mathrm{4}} \:+\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{3}\right){e}^{−{n}^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{n}^{\mathrm{2}} \:{U}_{{n}} =\frac{\pi}{\mathrm{16}{n}^{\mathrm{3}} }\left({n}^{\mathrm{4}} \:+\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{3}\right){e}^{−{n}^{\mathrm{2}} } \:\:\:=\frac{\pi}{\mathrm{16}}\left({n}\:+\mathrm{3}\:+\frac{\mathrm{3}}{{n}^{\mathrm{3}} }\right){e}^{−{n}^{\mathrm{2}} } \:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} {n}^{\mathrm{2}} \:{U}_{{n}} ={lim}_{{n}\rightarrow+\infty} \frac{\pi}{\mathrm{16}}\left({n}+\mathrm{3}\right){e}^{−{n}^{\mathrm{2}} } \:=\mathrm{0} \\ $$

Commented by maxmathsup by imad last updated on 16/Jun/19

2) we have nU_n =(π/(16n^4 ))(n^4 +3n^2 +3) e^(−n^2 ) ⇒lim_(n→+∞ ) nU_n = lim_(n→+∞) (π/(16)) e^(−n^2 ) =0

$$\left.\mathrm{2}\right)\:{we}\:{have}\:{nU}_{{n}} =\frac{\pi}{\mathrm{16}{n}^{\mathrm{4}} }\left({n}^{\mathrm{4}} \:+\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{3}\right)\:{e}^{−{n}^{\mathrm{2}} } \:\Rightarrow{lim}_{{n}\rightarrow+\infty\:} {nU}_{{n}} = \\ $$$${lim}_{{n}\rightarrow+\infty} \frac{\pi}{\mathrm{16}}\:{e}^{−{n}^{\mathrm{2}} } \:=\mathrm{0} \\ $$

Commented by maxmathsup by imad last updated on 16/Jun/19

4) the Q is study the convergence of Σ U_n

$$\left.\mathrm{4}\right)\:{the}\:{Q}\:{is}\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{U}_{{n}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com